Mixing
Show a finite topological space is path connected
$begingroup$
Topological space (X,T), where
$$X={v_1,v_2,e_1,e_2,u_1,u_2}$$
$$T={emptyset,{u_1},{u_2},{u_1,u_2},{e_1,u_1,u_2},{e_2,u_1,u_2},{e_1,e_2,u_1,u_2},{v_1,e_1,e_2,u_1,u_2},{v_2,e_1,e_2,u_1,u_2},X}$$
The question is to prove that the space is path connected. To show that, do I define a function from $[0,1]->X$ such that $f([0,1))=u_1$ and $f(1)=text{any other elements in }X$? And then show it is continuous? It doesn't seem right though.
I'm also not sure why the topology T is the way it is.
general-topology path-connected
edited Oct 31 '17 at 8:36
tommy xu3
9791621
asked Oct 31 '17 at 4:28
AreeddAreedd
826
$endgroup$
add a comment |
$begingroup$
Topological space (X,T), where
$$X={v_1,v_2,e_1,e_2,u_1,u_2}$$
$$T={emptyset,{u_1},{u_2},{u_1,u_2},{e_1,u_1,u_2},{e_2,u_1,u_2},{e_1,e_2,u_1,u_2},{v_1,e_1,e_2,u_1,u_2},{v_2,e_1,e_2,u_1,u_2},X}$$
The question is to prove that the space is path connected. To show that, do I define a function from $[0,1]->X$ such that $f([0,1))=u_1$ and $f(1)=text{any other elements in }X$? And then show it is continuous? It doesn't seem right though.
I'm also not sure why the topology T is the way it is.
general-topology path-connected
edited Oct 31 '17 at 8:36
tommy xu3
9791621
asked Oct 31 '17 at 4:28
AreeddAreedd
826
$endgroup$
$begingroup$
That is probably the best way to show a finite topological space is path connected.
$endgroup$
– Matt Samuel
Oct 31 '17 at 4:34
$begingroup$
If we do this , won't we have a problem? I mean, the preimage of a set which contains the second element but not the first, would just be ${1}$, and therefore, not an open set in $[0,1].$
$endgroup$
– астон вілла олоф мэллбэрг
Oct 31 '17 at 4:47
$begingroup$
Such a set need not be open.
$endgroup$
– Matt Samuel
Oct 31 '17 at 4:51
add a comment |
$begingroup$
Topological space (X,T), where
$$X={v_1,v_2,e_1,e_2,u_1,u_2}$$
$$T={emptyset,{u_1},{u_2},{u_1,u_2},{e_1,u_1,u_2},{e_2,u_1,u_2},{e_1,e_2,u_1,u_2},{v_1,e_1,e_2,u_1,u_2},{v_2,e_1,e_2,u_1,u_2},X}$$
The question is to prove that the space is path connected. To show that, do I define a function from $[0,1]->X$ such that $f([0,1))=u_1$ and $f(1)=text{any other elements in }X$? And then show it is continuous? It doesn't seem right though.
I'm also not sure why the topology T is the way it is.
general-topology path-connected
edited Oct 31 '17 at 8:36
tommy xu3
9791621
asked Oct 31 '17 at 4:28
AreeddAreedd
826
$endgroup$
Topological space (X,T), where
$$X={v_1,v_2,e_1,e_2,u_1,u_2}$$
$$T={emptyset,{u_1},{u_2},{u_1,u_2},{e_1,u_1,u_2},{e_2,u_1,u_2},{e_1,e_2,u_1,u_2},{v_1,e_1,e_2,u_1,u_2},{v_2,e_1,e_2,u_1,u_2},X}$$
The question is to prove that the space is path connected. To show that, do I define a function from $[0,1]->X$ such that $f([0,1))=u_1$ and $f(1)=text{any other elements in }X$? And then show it is continuous? It doesn't seem right though.
I'm also not sure why the topology T is the way it is.
general-topology path-connected
general-topology path-connected
edited Oct 31 '17 at 8:36
tommy xu3
9791621
asked Oct 31 '17 at 4:28
AreeddAreedd
826
edited Oct 31 '17 at 8:36
tommy xu3
9791621
asked Oct 31 '17 at 4:28
AreeddAreedd
826
edited Oct 31 '17 at 8:36
tommy xu3
9791621
edited Oct 31 '17 at 8:36
tommy xu3
9791621
edited Oct 31 '17 at 8:36
tommy xu3
9791621
9791621
asked Oct 31 '17 at 4:28
AreeddAreedd
826
asked Oct 31 '17 at 4:28
AreeddAreedd
826
asked Oct 31 '17 at 4:28
AreeddAreedd
826
826
$begingroup$
That is probably the best way to show a finite topological space is path connected.
$endgroup$
– Matt Samuel
Oct 31 '17 at 4:34
$begingroup$
If we do this , won't we have a problem? I mean, the preimage of a set which contains the second element but not the first, would just be ${1}$, and therefore, not an open set in $[0,1].$
$endgroup$
– астон вілла олоф мэллбэрг
Oct 31 '17 at 4:47
$begingroup$
Such a set need not be open.
$endgroup$
– Matt Samuel
Oct 31 '17 at 4:51
add a comment |
$begingroup$
That is probably the best way to show a finite topological space is path connected.
$endgroup$
– Matt Samuel
Oct 31 '17 at 4:34
$begingroup$
If we do this , won't we have a problem? I mean, the preimage of a set which contains the second element but not the first, would just be ${1}$, and therefore, not an open set in $[0,1].$
$endgroup$
– астон вілла олоф мэллбэрг
Oct 31 '17 at 4:47
$begingroup$
Such a set need not be open.
$endgroup$
– Matt Samuel
Oct 31 '17 at 4:51
$begingroup$
That is probably the best way to show a finite topological space is path connected.
$endgroup$
– Matt Samuel
Oct 31 '17 at 4:34
$begingroup$
That is probably the best way to show a finite topological space is path connected.
$endgroup$
– Matt Samuel
Oct 31 '17 at 4:34
$begingroup$
If we do this , won't we have a problem? I mean, the preimage of a set which contains the second element but not the first, would just be ${1}$, and therefore, not an open set in $[0,1].$
$endgroup$
– астон вілла олоф мэллбэрг
Oct 31 '17 at 4:47
$begingroup$
If we do this , won't we have a problem? I mean, the preimage of a set which contains the second element but not the first, would just be ${1}$, and therefore, not an open set in $[0,1].$
$endgroup$
– астон вілла олоф мэллбэрг
Oct 31 '17 at 4:47
$begingroup$
Such a set need not be open.
$endgroup$
– Matt Samuel
Oct 31 '17 at 4:51
$begingroup$
Such a set need not be open.
$endgroup$
– Matt Samuel
Oct 31 '17 at 4:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So, if we define $f([0,1))= u_1$ and $f(1) = y$, where $y neq u_2$, then note that $f^{-1}({u_1})$ is open, and any open set that contains $y$ also contains $u_1$, so for any such set $S$, $f^{-1}(S) = [0,1]$, which is open in $[0,1]$ obviously. Therefore, $f$ is continuous.
A similar logic applies for $u_2$ above.
So every point is path connected to $u_1$ and $u_2$, and you can now use the fact that paths can be reversed and concatenated to finish the argument. For example, to go from $v_1 to v_2$, first go to $u_1$ via the reverse of the path we made above, and then go to $v_2$ as usual.
(the next part is only for information. If you do not understand, it is fine)
I am inclined to feel that this topology comes from a topology we naturally establish on a poset, by considering chains. Here, we would have $u_i leq e_i leq v_i$ for each $i=1,2$, and the chains of this poset form a topology, that I think we call the Alexandroff topology on the poset. I would be inclined to think that the Alexandroff topology is path connected, under certain conditions on the poset.
Indeed, due to local path connectedness, one sees that the Alexandroff topology is path connected if and only if it is connected i.e. there is no partition of the set into disjoint open sets. Since every pair of non-empty open sets intersect at $u_1$ in this case, we see that the Alexandroff topology is connected, and therefore path connected without having to explicitly show the path.
answered Oct 31 '17 at 5:02
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
$endgroup$
$begingroup$
So we just need to show it again with $u2$, I see. Thank you so much!!!
$endgroup$
– Areedd
Nov 1 '17 at 13:44
$begingroup$
You are welcome. By the way, I was right about the Alexandroff topology. I think the Alexandroff topology is also "locally path connected", which means that you can find for every open neighbourhood around every pair of points, a path that stays in this neighbourhood, which connects the two points.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 2 '17 at 9:52
add a comment |
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$begingroup$
So, if we define $f([0,1))= u_1$ and $f(1) = y$, where $y neq u_2$, then note that $f^{-1}({u_1})$ is open, and any open set that contains $y$ also contains $u_1$, so for any such set $S$, $f^{-1}(S) = [0,1]$, which is open in $[0,1]$ obviously. Therefore, $f$ is continuous.
A similar logic applies for $u_2$ above.
So every point is path connected to $u_1$ and $u_2$, and you can now use the fact that paths can be reversed and concatenated to finish the argument. For example, to go from $v_1 to v_2$, first go to $u_1$ via the reverse of the path we made above, and then go to $v_2$ as usual.
(the next part is only for information. If you do not understand, it is fine)
I am inclined to feel that this topology comes from a topology we naturally establish on a poset, by considering chains. Here, we would have $u_i leq e_i leq v_i$ for each $i=1,2$, and the chains of this poset form a topology, that I think we call the Alexandroff topology on the poset. I would be inclined to think that the Alexandroff topology is path connected, under certain conditions on the poset.
Indeed, due to local path connectedness, one sees that the Alexandroff topology is path connected if and only if it is connected i.e. there is no partition of the set into disjoint open sets. Since every pair of non-empty open sets intersect at $u_1$ in this case, we see that the Alexandroff topology is connected, and therefore path connected without having to explicitly show the path.
answered Oct 31 '17 at 5:02
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
$endgroup$
$begingroup$
So we just need to show it again with $u2$, I see. Thank you so much!!!
$endgroup$
– Areedd
Nov 1 '17 at 13:44
$begingroup$
You are welcome. By the way, I was right about the Alexandroff topology. I think the Alexandroff topology is also "locally path connected", which means that you can find for every open neighbourhood around every pair of points, a path that stays in this neighbourhood, which connects the two points.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 2 '17 at 9:52
add a comment |
$begingroup$
So, if we define $f([0,1))= u_1$ and $f(1) = y$, where $y neq u_2$, then note that $f^{-1}({u_1})$ is open, and any open set that contains $y$ also contains $u_1$, so for any such set $S$, $f^{-1}(S) = [0,1]$, which is open in $[0,1]$ obviously. Therefore, $f$ is continuous.
A similar logic applies for $u_2$ above.
So every point is path connected to $u_1$ and $u_2$, and you can now use the fact that paths can be reversed and concatenated to finish the argument. For example, to go from $v_1 to v_2$, first go to $u_1$ via the reverse of the path we made above, and then go to $v_2$ as usual.
(the next part is only for information. If you do not understand, it is fine)
I am inclined to feel that this topology comes from a topology we naturally establish on a poset, by considering chains. Here, we would have $u_i leq e_i leq v_i$ for each $i=1,2$, and the chains of this poset form a topology, that I think we call the Alexandroff topology on the poset. I would be inclined to think that the Alexandroff topology is path connected, under certain conditions on the poset.
Indeed, due to local path connectedness, one sees that the Alexandroff topology is path connected if and only if it is connected i.e. there is no partition of the set into disjoint open sets. Since every pair of non-empty open sets intersect at $u_1$ in this case, we see that the Alexandroff topology is connected, and therefore path connected without having to explicitly show the path.
answered Oct 31 '17 at 5:02
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
$endgroup$
$begingroup$
So we just need to show it again with $u2$, I see. Thank you so much!!!
$endgroup$
– Areedd
Nov 1 '17 at 13:44
$begingroup$
You are welcome. By the way, I was right about the Alexandroff topology. I think the Alexandroff topology is also "locally path connected", which means that you can find for every open neighbourhood around every pair of points, a path that stays in this neighbourhood, which connects the two points.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 2 '17 at 9:52
add a comment |
$begingroup$
So, if we define $f([0,1))= u_1$ and $f(1) = y$, where $y neq u_2$, then note that $f^{-1}({u_1})$ is open, and any open set that contains $y$ also contains $u_1$, so for any such set $S$, $f^{-1}(S) = [0,1]$, which is open in $[0,1]$ obviously. Therefore, $f$ is continuous.
A similar logic applies for $u_2$ above.
So every point is path connected to $u_1$ and $u_2$, and you can now use the fact that paths can be reversed and concatenated to finish the argument. For example, to go from $v_1 to v_2$, first go to $u_1$ via the reverse of the path we made above, and then go to $v_2$ as usual.
(the next part is only for information. If you do not understand, it is fine)
I am inclined to feel that this topology comes from a topology we naturally establish on a poset, by considering chains. Here, we would have $u_i leq e_i leq v_i$ for each $i=1,2$, and the chains of this poset form a topology, that I think we call the Alexandroff topology on the poset. I would be inclined to think that the Alexandroff topology is path connected, under certain conditions on the poset.
Indeed, due to local path connectedness, one sees that the Alexandroff topology is path connected if and only if it is connected i.e. there is no partition of the set into disjoint open sets. Since every pair of non-empty open sets intersect at $u_1$ in this case, we see that the Alexandroff topology is connected, and therefore path connected without having to explicitly show the path.
answered Oct 31 '17 at 5:02
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
$endgroup$
So, if we define $f([0,1))= u_1$ and $f(1) = y$, where $y neq u_2$, then note that $f^{-1}({u_1})$ is open, and any open set that contains $y$ also contains $u_1$, so for any such set $S$, $f^{-1}(S) = [0,1]$, which is open in $[0,1]$ obviously. Therefore, $f$ is continuous.
A similar logic applies for $u_2$ above.
So every point is path connected to $u_1$ and $u_2$, and you can now use the fact that paths can be reversed and concatenated to finish the argument. For example, to go from $v_1 to v_2$, first go to $u_1$ via the reverse of the path we made above, and then go to $v_2$ as usual.
(the next part is only for information. If you do not understand, it is fine)
I am inclined to feel that this topology comes from a topology we naturally establish on a poset, by considering chains. Here, we would have $u_i leq e_i leq v_i$ for each $i=1,2$, and the chains of this poset form a topology, that I think we call the Alexandroff topology on the poset. I would be inclined to think that the Alexandroff topology is path connected, under certain conditions on the poset.
Indeed, due to local path connectedness, one sees that the Alexandroff topology is path connected if and only if it is connected i.e. there is no partition of the set into disjoint open sets. Since every pair of non-empty open sets intersect at $u_1$ in this case, we see that the Alexandroff topology is connected, and therefore path connected without having to explicitly show the path.
answered Oct 31 '17 at 5:02
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
edited Jan 18 at 6:26
answered Oct 31 '17 at 5:02
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
answered Oct 31 '17 at 5:02
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
answered Oct 31 '17 at 5:02
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.8k33477
38.8k33477
$begingroup$
So we just need to show it again with $u2$, I see. Thank you so much!!!
$endgroup$
– Areedd
Nov 1 '17 at 13:44
$begingroup$
You are welcome. By the way, I was right about the Alexandroff topology. I think the Alexandroff topology is also "locally path connected", which means that you can find for every open neighbourhood around every pair of points, a path that stays in this neighbourhood, which connects the two points.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 2 '17 at 9:52
add a comment |
$begingroup$
So we just need to show it again with $u2$, I see. Thank you so much!!!
$endgroup$
– Areedd
Nov 1 '17 at 13:44
$begingroup$
You are welcome. By the way, I was right about the Alexandroff topology. I think the Alexandroff topology is also "locally path connected", which means that you can find for every open neighbourhood around every pair of points, a path that stays in this neighbourhood, which connects the two points.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 2 '17 at 9:52
$begingroup$
So we just need to show it again with $u2$, I see. Thank you so much!!!
$endgroup$
– Areedd
Nov 1 '17 at 13:44
$begingroup$
So we just need to show it again with $u2$, I see. Thank you so much!!!
$endgroup$
– Areedd
Nov 1 '17 at 13:44
$begingroup$
You are welcome. By the way, I was right about the Alexandroff topology. I think the Alexandroff topology is also "locally path connected", which means that you can find for every open neighbourhood around every pair of points, a path that stays in this neighbourhood, which connects the two points.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 2 '17 at 9:52
$begingroup$
You are welcome. By the way, I was right about the Alexandroff topology. I think the Alexandroff topology is also "locally path connected", which means that you can find for every open neighbourhood around every pair of points, a path that stays in this neighbourhood, which connects the two points.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 2 '17 at 9:52
add a comment |
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$begingroup$
That is probably the best way to show a finite topological space is path connected.
$endgroup$
– Matt Samuel
Oct 31 '17 at 4:34
$begingroup$
If we do this , won't we have a problem? I mean, the preimage of a set which contains the second element but not the first, would just be ${1}$, and therefore, not an open set in $[0,1].$
$endgroup$
– астон вілла олоф мэллбэрг
Oct 31 '17 at 4:47
$begingroup$
Such a set need not be open.
$endgroup$
– Matt Samuel
Oct 31 '17 at 4:51